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Let $k:X \to Y$ be an onto map.How to prove that the quotient topology on $Y$ induced by $k$ is the largest topology relative to which $k$ is continuous.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. –  Julian Kuelshammer Oct 29 '12 at 12:55
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What is your definition of the quotient topology? –  commenter Oct 29 '12 at 12:59
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To add to what Julian said: this is not the first question you post so you should learn how to display latex, namely by wrapping all your maths in dollar signs. –  Matt N. Oct 29 '12 at 13:03
    
@ commenter,If a map p:X to Y is surjective,continuous and either open or closed then the map p is a quotient map.then the topology induces by p is the quotient topology.that's what I know.But I have no practice with doing problems in this.so can you help me with this. –  ccc Oct 29 '12 at 13:05
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But what is the quotient topology, according to you? It might help if you spelled out all the definitions in detail. What is the quotient topology induced by k and what is the largest topology with respect to which k is continuous? Write them down and then it might be possible to give an answer different from "by definition". –  commenter Oct 29 '12 at 13:08
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up vote 3 down vote accepted

I’m going to assume that you’ve defined the quotient topology $\tau$ on $Y$ as follows: a set $U\subseteq Y$ is open iff $k^{-1}[U]$ is open in $X$. It’s an immediate consequence of this definition that the topology $\tau$ on $Y$ makes $k$ continuous, so what you have to show is that if $\tau'$ is another topology on $Y$ making $k$ continuous, and $\tau\subseteq\tau'$, then $\tau=\tau'$. In other words, you must show that it’s impossible to have $\tau'\supsetneqq\tau$: there is no topology on $Y$ that is strictly stronger than $\tau$ and makes $k$ continuous.

Here’s a large hint to get you started. Suppose that $\tau\subseteq\tau'$, where $\tau'$ is a topology on $Y$ making $k$ continuous. If $\tau'\ne\tau$, there is a set $U\in\tau'\setminus\tau$. By hypothesis $k$ is continuous as a map from $X$ to $\langle Y,\tau'\rangle$, so $k^{-1}[U]$ is open in $X$. Now apply the definition of the quotient topology $\tau$ to get a contradiction.

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@ Brian,Thank you very much for your hint..now I understood the problem and the way of getting the answer. –  ccc Oct 30 '12 at 8:02
    
@cccjay: You’re welcome; glad it helped. –  Brian M. Scott Oct 30 '12 at 14:14
    
@ Brian,I clearly understood the idea here.but how can I relate the condition that the map k is surjective to get this contradiction? –  ccc Nov 8 '12 at 7:00
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@ccc: Recall that a set $V\subseteq Y$ is in $\tau$ iff $k^{-1}[V]$ is open in $X$, by the definition of the quotient topology $\tau$. And $k^{-1}[U]$ must be open in $X$, since $k$ is continuous with respect to $\tau'$. But then $U\in\tau$, contrary to our choice of $U$. –  Brian M. Scott Nov 8 '12 at 8:05
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